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					                   SIGHT REDUCTION FOR NAVIGATION

                                       Table of Contents

I. Time

       A. Time Itself                                       2

       B. Time and the Noon Sight

              1. Latitude by Mer Pass at Lan                3

              2. Longitude by Mer Pass, Sunrise or Sunset   3

II. The Navigational Triangle

       A. Navigational Triangle                             4

       B. Oblique Spherical Triangle                        5

       C. Right Spherical Triangles                         8

              1. Ageton H.O. 211                            8

              2. Dreisonstok H.O. 208                       9

       D. Comparison of Ageton, Dreisonstok, and H.O. 9     10

              1. Definitions                                10

              2. Mathematics                                11

III. The Great Circle Track                                 13

IV. Useful References                                       14



Time is longitude. Geocentrically speaking the sun goes round the earth 360 degrees every twenty
four hours, more or less. That is, spherically:

ANGLE                 TIME                                 DISTANCE               DISTANCE
                                                           Equator                30 North Lat

360°   equals         24 hours              equals         21,600 NM              14,400 NM
15°    equals          1 hour               equals            900                    600
1°     equals          4 minutes            equals             60                     40
15N    equals          1 minute             equals             15                     10
1N     equals          4 seconds            equals              1                  1,235 m
15O    equals          1 second             equals          1,235 m                  309 m

 Since Greenwich is 0° longitude (symbol lambda 8) and Dallas is 8 96° 48N W, Dallas is 6 hours,
27 minutes, 12 seconds west of Greenwich in time. This is exact time from place to place so noon
local mean time would be 18h27m12s GMT..

Once upon a time there was much ado about time zones, i.e, Central or Pacific, and about watch or
chronometer error, but now since everyone has reference to GMT by his computer, watch or radio,
GMT (recently officially called UTC) is the gold standard. You can even set it as your home page
on the web by going to the USNO/NIST site Local time and zone time can
generally be disregarded.



Local Apparent Noon (LAN) is an event, not a time system. You can determine LAN by looking
at your GMT watch and timing successive observations around local noon or by calculation if you
know the longitude. In Dallas, this would, by calculation, be at GMT 18h27m12s, but there is a
catch. GMT means Greenwich Mean Time. It is mean time because it is averaged time. Depending
on the time of year the actual Meridian Passage (Mer Pass) of the sun can be 16m25s earlier than the
GMT or14m15s later than GMT depending upon “the obliquity of the ecliptic”.

These values are called the equation of time (Eq.T) and must be added or subtracted from your
calculated GMT to arrive at the exact time of Mer Pass or LAN expressed in GMT. The values of
the Eq. T can be found in the Nautical Almanac or on the Web at
bin/ Be careful where you get the values for Eq.T: the sign + or - is an integral part
of the value. Go to USNO for the US navigation values. In the US, the formula for apparent and
mean time is: Apparent time equals Mean time minus Eq.T. Thus if Eq.T is positive, the time value
is subtracted from the calculated GMT to arrive at LAN; if the Eq.T is negative, it is added to
calculated GMT to arrive at LAN. The rule is backwards in more backward parts of the world like
UK. Thus in Dallas, the calculated GMT for 0 Eq.T is GMT 18h 27m12s, but in February, Mer Pass
is as late as 18h41m26s, and in November, it is as early as 18h 10m50s.

Graphically you can see both the Eq.T and declination (d) on an analemma- the strange figure eight
on a globe. Different websites draw different analemmas and use different signs. However it is
drawn, the May and November side is a +Eq.T and is subtracted from GMT to get LAN. The August
and March side is -Eq. T and is added to GMT to determine the time of LAN.

The noon sunshot is acknowledged as the most reliable LOP of the day and is important to
determining latitude because there is no sight reduction involved except addition and subtraction:
90° plus or minus declination minus Ho (observed altitude) equals latitude.


It is possible to determine longitude from the noon sunshot but you may well be inaccurate by a
degree either way since the sun moves almost horizontally at noon and determining the exact time
of LAN by a series of sunshots gives only a rough fix in about a three or four minute window.
Assume that you have determined the GMT of LAN by observation. In order to use the
time=longitude formula (one hour equals 15 degrees) you must also determine the GMT of LAN at
Greenwich. Otherwise your measurement will be elongated or shorted by the Eq.T. So, if on July
8 the Eq. T is approximately -5m the LAN in Greenwich will be about 1205 GMT and the LAN in
Dallas will also be 5 minutes later than expected, probably about 18h32. Remember that LAN is an
event not a time system. To properly measure the time and distance on the earth’s surface remember
that we have to measure the GMT of the LAN event at both ends of the calculation. You can also
watch sunrise and sunset but the exact timing requires careful corrections.

                             THE NAVIGATIONAL TRIANGLE

Spherical trigonometry is rarely encountered in our times. The main practical use was air and sea
navigation. The methods for sight reduction dealt with the “oblique” and “right” spherical triangles.
The mathematical theory wasn’t hard, but the tedious arithmetic was a killer. Logarithmic functions
at least reduced the arithmetic to adding and subtracting. Today’s hand calculator makes it easy.


Consider the navigational spherical triangle: The vertices are the pole (P), your assumed position
(AP, also called Z), and the geographic position (GP, also called M) of the celestial body. You
know two of the sides because from declination, you can determine its complementary polar distance
or “co-d”; and, from estimated latitude, you can determine its complementary “co-Lat”. What we
are trying to learn is the third side between AP and GP since that side, known as “co-H”, is the
complement of Hc. The prerequisites are knowning LHA or t, declination or d of the celestial body,
and assumed position AP. You will calculate LHA and learn d from the almanac; AP is your
assumed position based on your dead reckoning. GHA Aries + SHA of the celestial body = LHA
+ Longitude.

Here are four formulas for solving the oblique spherical triangle without dividing it into two
spherical right triangles. The first H.O. 9 formula is the root of all the rest. You can look up sin L
and cos L, and sin d and cos d, together since the values usually appear across the page from each
other in a book of trigonometric values. These are all variants of the law of cosines for sides.

1. H.O. 9 (1966 ed) page 528:

        sin h = sin L sin d + cos L cos d cos t

        which can be transformed into:

        sin h = sin L sin d + sin (co-Lat) sin (co-d) cos t

2. Dutton (15th ed.) par. 2502:

        sin H = sin L sin d + cos L cos d cos t

3. Nautical Almanac 2007 page 279:

        Hc = sin-1 (S sin Lat + C cos Lat)

        which transforms into:

        sin Hc = (sin d sin L + cos d cos t cos L)

4. Hobbs (4th ed. 1998) page 411, uses the NAO notation which turns out to be:

        sin Hc = (Sin L sin d + cos L cos d cos t)

These are all derived from the law of cosines. See Frank Ayers, Plane and Spherical Trigonometry
(1st ed. 1954) page168.

The cosine of the side ( side a) opposite a known angle A is:

        cos a = cos b cos c + sin b sin c cos A.

If we re-label this formula in nautical terms, it is:

        cos (co-H) = cos (co-d) cos (co-L) + sin (co-d) sin (co-L) cos t

Now transform:

        sin H = sin d sin L + cos d cos L cos t

which is the basic formula used by H.O. 9. Obviously you know d and t from the Almanac, and
have assumed aL, so you can do this with a hand calculator with ease.

An Example of solving the Oblique Spherical Triangle by Trigonometry:

Local time:              10:00 p.m., 31 Dec 2006
Place:                   USNA, Lat 38 59; Long 76 29
UTC:                     0300h 01 Jan 2007
Celestial body:          Betelguese
GHA                      56° 29.9'
Dec                      N 7° 24.6'
Hc                       53° 39.8'
Zn                       145.1°

The basics of the oblique spherical triangle “LTD”:

       L:         aLat = 38° 59'; co-lat = 51° 01'

       T:         Polar vertex angle, t, is determined by subtracting GHA of the body from longitude:
                  76° 29' minus 56° 30' = 19° 59'

       D:         Declination = 7° 24.6'; co-declination = 82° 35.4'

Using the H.O. 9 formula:

sin h = sin L sin d + cos L cos d cos t:

       sin h = sin L (38° 59) sin d (7° 24.6) + cos L (38° 59) cos d (7° 24.6) cos t (19° 59)

       sin h = (.62909) (.12908) + (.77715) (.99163) (.93979)

       sin h = (.0812029) + (.7242446)

       sin h = .8054475

       H = 53° 39'

       Co-h = 36° 21'

The azimuth of the celestial body can be determined by the law of sines.

       Sin A =           Sin B =         Sin C
       sin a             sin b           sin c

We have just determined that t = 19° 59' and co-h = 36° 21' so the ratio has been established.
Likewise we know that the side opposite vertex AP is the co-declination of 82° 35.4'. Thus the set
up to be solved is:

       sin t             =       sin AP
       sin co-h                  sin co-d

       sin 19° 59     =       sin AP
       sin 36° 21             sin 82° 35

       .34202         =       sin AP
       .59272                 .99163

       sin AP         =       .57220

       AP             =       34° 54'

This can’t be true! Betelguese is in the southeast. Remember that sines go from 0 at 0° to 1 at 90°
and back down to 0 at 180°. Thus sin AP at .57220 could be both 34° 54' or 145° 06'. Sin AP = Sin
(180°-AP). In this case it is clear that the angle at vertex AP is 145° 06'.


1. ARTHUR AGETON’S H.O. 211 SOLUTION. Ageton’s solution to this spherical triangle is to
divide it into two right spherical triangles by dropping a perpendicular (H) from GP to the opposite
side (the longitude line including AP to Pn). This method was used by Arthur Ageton in 1931 and
is the basis for H.O. 211. The formulas are straight-forward except they involve a lot of substitutions
of co-functions of the complement of an angle for the function itself of that angle; i.e., sin 30° = cos
60°; sec d = csc (90°-d). In H.O. 211 Ageton uses log secants (the B column) and log cosecants (the
A column). Remember co-d = 90-d; co-lat =90-lat. If h is the height of the celestial body, then co-h
has to be the side of the triangle between GP and AP.

There are three calculations to determine Hc in the Ageton system:

1. The first calculation is determination of R which is the side between M and the right angle
intersection of the perpendicular to side Pn-AP (or the extension of Pn -AP) at point X. We know
two angles: the 90° degree angle at vertex X, and t or LHA at Pn; we also know side co-d as the
complement of the declination of the heavenly body.

The formula would be:          csc R = csc t C csc (co-d)

Ageton inverts csc (co-d) into sec d, and comes up with a formula:

                               csc R = csc t C sec d

Since Ageton uses logarithms, his formula is:

                               log csc R = log csc t + log sec d

2. The second calculation is determination of side K from the equator to X which is based on
calculating co-K from Pn to X. Co-K is easily enough calculated because we know sides co-d and
R. The fundamental formula is:

                               sec co-K = csc d / sec R

Ageton transforms this into: csc k = csc d / sec R

or, in logarithmic terms:      log csc K = log csc d - log sec R

At this point subtract aLat from K if aLat is smaller and of the same sign or add aLat to K if they
are of different signs. This produces a value which Ageton calls K~L and which is sometimes called

The Bayless Compact Sight Reduction Table (Modified H.O. 211) In 1980 Allan Bayless published
a nine-page shortened version of the Ageton table of log secants and log cosecants. The entire
booklet is 32 pages long and an easier presentation of Ageton’s format and tables.

2. DREISONSTOK’S H.O. 208 SOLUTION Alternatively one can drop the perpendicular from
the AP vertex to the co-d side as is done in the NAO Concise Sight Reduction Table or by
Dreisonstok’s H.O. 208.


    1. Definitions. Both Dreisonstock and Ageton solve the oblique spherical triangle by dividing it into
    two right spherical triangles. Ageton drops a perpendicular from the GP of the celestial body to the
    P-AP line; Dreisonstock drops a perpendicular from the AP to the P-GP line. Some terminology is
    the same; some is different.

             AGETON                            DREISONSTOK                             BOWDITCH
             H.O. 211                             H.O. 208                             H.O. 9 (1966)

LHA =t                                LHA =t                                  LHA =t

P                                     P                                       Pn

Z = AP                                Z = AP                                  Z= AP

M = GP                                M = GP                                  M= GP

Co-L = colatitude                     Co-L = colatitude                       Co-L = colatitude

co-d = codeclination                  co-d = codeclination                    Co-d = codeclination

co-h = co-altitude side Z-M           co-h = co-altitude side Z-M             co-h = co-altitude side Z-M

R = the perpendicular from M          a = the perpendicular from Z            v = the perpendicular from Z
intersecting at X on side P-Z; v; a   intersecting at D on side P-M; v; r     intersecting side P-M

X = 90 intersection on side Pn-Z      D= 90 intersection on side Pn-M         w

w = part side P-Z between P and       b = part side P-M between P and         x
X (90 -K); comparable to b            D; comparable to w
                                                                              See figures 2110 and 2111.
x = K minus Lat; Lat minus K;
K~L                                   B = part side P-M between M and
K = equinoctial to X

co-K = X to P; w

L = equinoctial to Z

2. Comparison of the Mathematics of Ageton and Dreisonstok

 AGETON                                             DREISONSTOK
 H.O. 211                                           H.O. 208

 First Calculation:                                 First Calculation:

 Determination of right side R which is the         Determination of right side a which is the
 side opposite t between M (GP) and the right       side opposite t between Z (AP) and the right
 angle at point X. We know two angles; the          angle at point D. We know two angles: the
 90° degree angle at vertex X and t (or LHA)        90° degree angle at point D and t (or LHA) at
 at P; we also know hypotenuse side co-d as         P; we also know hypotenuse side co-L as the
 the complement of declination d.                   complement of latitude L.
 Remember a = R                                     Remember a = R

 Formula:                                           Formula:
                                                    sin a = sin t C cos L
                                                    sin a = sin t C sin (co-L)
 csc R = csc t C Csc (co-d)                         csc a = csc t C csc (co-L)
 csc R = csc t C sec d                              csc a = csc t C sec L

 log csc R = log csc t + log sec d

 Second Calculation:
                                                    Second Calculation:
 Determination of right side K from the
 equator to X which is based on calculating         Determination of right side B
 co-K from P to X. Co-K is easily enough
 calculated because we know sides co-d and R.       Formula:

 Formula:                                           tan b = cot L cos t

 sec co-K = csc d / sec R                           B = 90 - d - b
 csc K = csc d / sec R
 log csc K = log csc d - log sec R                  B = 90 - (d+b)

 At this point subtract a Lat from K if aLat is
 smaller and of the same sign or add a Lat to K
 if they are of different signs. This produces
 as value which Ageton calls K~L and which
 is sometimes called x.

Third Calculation.                           Third Calculation:

Determination of h (altitude) or co-h        Determination of h (altitude) or co-h
(hypotenuse side between AP and GP) when     (hypotenuse side between Z and M) when
right sides R and K~L are known.             right sides a and B are known.

                                             90° = B + b +d

Formula:                                     Formula:

sec co-h (side) = sec R C Sec x              sin h (alt) = cos a C cos B

                                             sin h (alt) = cos a C sin (d+b)

                                             cos co-h (side) = cos a C cos B

                                             sin h = cos a C sin B

                                             csc h (alt) = sec a C csc (d +b)

csc h (alt) = sec R C sec x                  csc h(alt) = sec a C sec B

remember (a = R) and (x = B)                 remember (a = R) and (x = B)

                              THE GREAT CIRCLE TRACK

Exactly the same formulas for an oblique spherical triangle which apply to determining Hc apply to
determining the mileage of the great circle track between two ports (after all, what is co-H except
the great circle between AP and GP). The polar angle t is known; the two adjacent sides, colatitudes
are calculated by subtracting the latitude from 90°.

St. John’s Newfoundland is at 47° 34' N 52° 42' W; its co-latitude is 42° 26'.

Kinsale Light Old Head is at 51° 36' N 8° 32' W; its co-latitude is 38° 28'.

The polar angle (known as t ) between these two longitudes is 44° 10', cos A = .71732.

                        Co-Lat                  Cos                      Sin

St. Johns               42° 26'                 .73806                   .67473

Kinsale                 38° 28'                 .78297                   .62092

Using the cosine formula for two sides and an included angle, we have:

cos a = cos b cos c + sin b sin c cos A

cos a = (.73806) (.78297) + (.67473) (.62206) (.71732)

cos a = (.577878) + (.301075)

cos a = 0.87895

a = 28° 29' x 60 nm/degree

a = 1707 nm

You might initially think that it would be easier to work out a right triangle with longitude and
latitude lines but recall that, although all longitudes are great circles, the only latitude line which is
a great circle is the equator.

With these formulas it is easy to determine the course at beginning and at end, but the proper course
throughout the voyage is always changing. Use an AP for the beginning of the voyage and

                                   USEFUL REFERENCES

The website is the Astronomical Applications Department of the United
States Naval Observatory. It carries on the web the calculations that one would make from the
Nautical Almanac and H.O. 229 (which has replaced H.O. 214). The Eq.T for any particular time
is at, and a complete set of GHA, declination, and
Hc for selected bodies at the chosen AP is at

Spherical trigonometry is treated in Frank Ayers, Jr., Plane and Spherical Trigonometry (Schaums’s
Outline Series, 1st ed. 1954), and in Kells, Kern & Bland, Spherical Trigonometry with Naval and
Military Applications (USNA, 1942).


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